• Functional anatomy• Tissue mechanics: mechanical properties of

muscle, tendon, ligament, bone• Kinematics: quantification of motion, with no

regard for the forces• Kinetics: forces, torques

Overall course plan

Reading

• Forces & COM:– Ch 2: 41-42, 46-54; Ch 3: 115-117

• Ground Reaction Forces & Locomotion:– Ch 2: 56-59

• Pressure, COP, Friction:– Ch 2: 60-63

• Free-Body Diagrams– Ch2:43-44; Ch 3:107-109

• How do we propel ourselves forward during running?

• Why does a runner lean on the curve in a track?

• What keeps an airplane in the air?• How does a pitcher throw a curve ball?

What is a force?

• “an agent that produces or tends to produce a change in the state of rest or motion of an object”

• Kinetics = study of motion resulting from forces• Forces are vectors:

– Magnitude– Direction

Force is a vector

What are the horizontal and vertical components of a force with a magnitude of 72 N acting at 16 degrees above the horizontal?

Outline• Kinetics: Forces in human motion

• Newton’s Laws– Law of Gravitation

– Center of Mass (C.O.M.)– Laws of motion:

– First Law: Inertia – Second Law: Acceleration– Third Law: Action-Reaction

• Pressure• Friction

Law of Gravitation

• All objects with mass attract one another.• Earth’s Grav. Force on a human = mg = weight• Unit of measurement of force = Newton (N)

Newton = kg • m / s2

• Human: weight of 154 lbs.– mass of 70 kg– weight of 700 N

• 6 fig newtons = 1 Newton• One small apple ~ 1 Newton

Center of mass (C.O.M.)

• A point about which all of the mass of an object is evenly distributed.

• Whole body can be represented by single point mass at C.O.M.

Position of C.O.M. in common shapes of uniform density

In the middle

of a cube

In the middleof a circle or

sphere

In the middle

of a square

C.O.M. in standing person

A crude method for finding C.O.M.

When balanced,the C.O.M. position

been found

Tips down

C.O.M. position can move.

C.O.M. of whole body: depends on the configuration of the body segments.

From Enoka 2.11

C.O.M. position depends on the distribution of body weight among and within the body

segments• Segmental analysis– head– trunk– upper arm– fore arm– hand– thigh– shank– foot

Enoka 2.13

Table 2.1

Segment mass

• Table 2.1 in Enoka - weights of segments and the position of the C.O.M. of segments.

• Example: thigh Thigh weight = 0.127•BW - 14.8 Thigh weight and BW are in Newtons

If person weighs 700N, thenThigh weight = 74.1 N

Where is the C.O.M. of the

thigh?

• Proximal: Closer to trunk.

• Distal: Further from trunk

Proximal

Distal

Thigh

C.O.M. is ~ 40% ofthe distance from theproximal to thedistal end of thigh

C.O.M. can be located outside the body

Enoka 2.11

C.O.M. and the brain

Outline• Kinetics: Forces in human motion

• Newton’s Laws– Law of Gravitation

– Center of Mass (C.O.M.)– Laws of motion:

– First Law: Inertia – Second Law: Acceleration– Third Law: Action-Reaction

• Pressure• Friction

Newton’s 1st Law:Law of Inertia

• Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by a force impressed upon it– Inertia = resistance of an object to motion• Directly related to body mass

• If SF=0, then Dv=0

Outline• Kinetics: Forces in human motion

• Newton’s Laws– Law of Gravitation

– Center of Mass (C.O.M.)– Laws of motion:

– First Law: Inertia – Second Law: Acceleration– Third Law: Action-Reaction

• Pressure• Friction

Newtons 2nd law:Law of acceleration

A force applied to a body causes an acceleration of a magnitude proportional to the force, in the direction of the force, and inversely proportional to the body’s mass.

SF = ma

SFx = max

SFy = may

a = 5 m/s2

m = 50 kgF = ma = 250 N

a = 5 m/s2

m = 100 kgF = ma = 500 N

250 N 500 N

How much force must be exerted to accelerate a 240kg mass to 5.7m/s2?

A)1368 kg m/s2 B) 1368 NC) 1368 kgD)890 m/s2

E) none of the above

Force in angular motion

• F = ma• Tangential

at = ∆v / ∆t Ft = mat = m∆v/∆t

• Radial ar = v2 / r Fc = mv2 / r centripetal force

Ft = m∆v/∆t

Fc = mv2/r

Force in angular motion: F = ma

ar = r2

at = r

Fc = mr2

Ft = mr

A 70 kg person is running around a circular path (radius = 2 m) with a constant angular velocity of 1 rad /sec. What are the magnitudes

of the average tangential and radial forces are required for this turn?

A)Ft= 140 N, Fc= 0 NB)Ft= 0 N, Fc= 140 rad/s2

C)Ft= 0 N, Fc= 140 ND)Ft= 140 N, Fc= 0 rad/s2

Jeff Francis pitches a 250 g baseball with a release velocity of 100MPH (44m/sec). His arm is fully extended at release, and the distance from the ball to his shoulder is 60cm. How much force is needed at the shoulder to keep his arm in place?

A) 806675 NB) 806.7 NC)10.995 ND)None of the above

Outline• Kinetics: Forces in human motion

• Newton’s Laws– Law of Gravitation

– Center of Mass (C.O.M.)– Laws of motion:

– First Law: Inertia – Second Law: Acceleration– Third Law: Action-Reaction

• Pressure• Friction

Law of action-reaction

• If one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object.

Example of action-reaction• A person exerts a force on

the floor equal to their body weight (mg).

• The floor exerts an equal and opposite force (mg) on the person.– “Ground reaction force”

(GRF)

Ground reaction force

mg

mg

Free body diagram recipe:• Simple sketch• Outline the defined “system” with ------• Arrow for force of gravity acting on c.o.m.• Arrows for GRFs• Arrows for other external forces• No arrows for internal forces• See Ch 2 page 43

Measurement of ground reaction force

• Static: a bathroom scale can be used.

• Dynamic (e.g., jump): a force platform should be used.

Force platform• Can measure forces very quickly (e.g., every

0.0001 s).• Can measure the forces in three dimensions– Vertical: Fg,y

– Horizontal: Fg,x

– Lateral: Fg,z

Y

X

Z

Sign convention for Fg,y

• Fg,y

Upward is positiveLocomotion, jumping, etc.: Fg,y > 0

Fg,y < 0: not normally possible need suction cup shoes

+Fg,y

Sign conventions

• Positive Fg,x : oriented in the direction of motion or toward anterior side.

+Fg,x

Analysis of ground reaction forces

• Can be used to calculate the acceleration of the C.O.M.

SF= ma

SFy= may

Fg,y - mg = may

Fg,y = may + mg

Ground reaction force

mg

mg

Analysis of ground reaction forces• Can be used to calculate the acceleration of the

C.O.M.

SF= ma

SFx= max

Fg,x = max

+Fg,x

Analysis of ground reaction forces

• Can be used to calculate the acceleration of the C.O.M.

SF= ma

SFy= may SFx= max

Fg,y - mg = may Fg,x = max

Fg,y = may + mg

Vertical jump

BodyweightF g,y (

N)

Fg,y = may + mg

RUN3.9 m/s

2100

1400

700

00 0.25 0.50 0.75

Time (s)

F g,y (

N)

RUN

F g,y

Stance

RUN3.9 m/s

500

0

-500

F g,x (

N)

0 0.25 0.50 0.75

Time (s)

Backward

Forward

Time (s)

WALK1.25 m/s

0

700Fg,y (N)

0 0.4 0.8 1.2

350

1050

-210

0

0 0.4 0.8 1.2Time (s)

Fg,x (N)

WALK1.25 m/s

Backward

Forward

210

Walk: 1.25 m/s Run: 3.9 m/s

x

PeakFg,y

(kN)

Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase:

Fg,x = -286NFg,y = 812NFg,z = 61N

Calculate the magnitude and direction for each of the resultant ground reaction force in the sagittal and frontal planes.

Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase:

Fg,x=-286NFg,y=812NFg,z=61N

Calculate the magnitude and direction for each of these resultant ground reaction force in the sagittal and frontal planes.

A) 861N at 0.34 rad relative to the vertical 814 N at 0.07 rad relative to the verticalB) 861N at 2.8 rad relative to the horizontal 814 N at 3.07 rad relative to the horizontalC) 861N at 19.4 rad relative to the vertical 814 N at 4.01 rad relative to the verticalD) 861N at 70.6 rad relative to the horizontal 814 N at 85.99 rad relative to the horizontalE) None of the above

Suppose a runner experienced the following ground reaction forces at one point in time during the stance phase:

Fx=-286NFy=812NFz=61N

Calculate the magnitude and direction of the resultant ground reaction force in the transverse plane.

A) 292N at 0.21 rad relative to the forward horizontal

B) 292N at 0.21 rad relative to the backward horizontal

C) 861N at 0.34 rad relative to the forward horizontal

D) 861N at 0.34 rad relative to the forward horizontal

Dynamic and Static Analyses

Dynamic AnalysisSF= ma

Static AnalysisSF= 0

Outline• Kinetics: Forces in human motion

• Newton’s Laws– Law of Gravitation

– Center of Mass (C.O.M.)– Laws of motion:

– First Law: Inertia – Second Law: Acceleration– Third Law: Action-Reaction

• Pressure• Friction

Force versus pressure

• Force: total force measured in Newtons• Pressure = force per unit area.– Force is measured in Newtons (N).– Area is measured in m2

– Pressure: N / m2 or Pascal (Pa)– P=F/A

Example of Fg versus pressure• Running shoe– Fg = 2000N.– Sole area = 0.025 m2.– Average pressure on sole of shoe (F / A)• F / A = 2000 N / 0.025 m2 = 80,000 N/m2

• Cleats (e.g., for soccer)– Fg = 2000N.– Cleat total area = 0.005 m2.– Average pressure on cleats (F / A)• F / A = 2000 N / 0.005 m2 = 400,000 N/m2

Distribution of pressure under sole of foot

• Divide sole into many small squares• Find pressure for square (F/A)– Use special insoles or pressure sensing mats to do

this

Distribution of pressure under sole of foot• Divide sole into many small squares• Find pressure for square (F/A)– Use special insoles or pressure sensing mats to do

this

– Center of Pressure:• Weighted average of all the downward acting forces• indicates the path of the resultant ground reaction force

vector

Rearfootstriker

Midfootstriker

Point of forceapplication(Centerof Pressure) Foot strike

Toe off

Enoka 2.19

The center of pressure and point of peak pressure are the same

A) TrueB) FalseC) It depends

Possibly confusing terms• Force• Pressure• Center of Pressure• Peak pressure• Point of force application

Outline• Kinetics: Forces in human motion

• Newton’s Laws– Law of Gravitation

– Center of Mass (C.O.M.)– Laws of motion:

– First Law: Inertia – Second Law: Acceleration– Third Law: Action-Reaction

• Pressure• Friction

Friction• Resistance to one object sliding, rolling, or flowing over another

object or surface.• Human movement examples: – foot - ground.– bicycle tire - ground– snowboard on snow– friction in joints

• Friction is a reaction force

Fa,x

Fs = friction force

Static Friction force (Fs)

• Friction force depends on:– properties of the surfaces– force acting perpendicular to the surfaces• e.g., ground-foot friction: Fg,y is the perpendicular force

Data from Miller and Nelson 1973.

Figure 2.18

Fs,max= * Fg,y

• Fg,y is vertical ground reaction force

• is coefficient of friction– dimensionless– depends on the properties of

the surfaces– static (s)

dynamic (d) (s > d)

Fa,x

Fg,y

Fs

Fa,x = applied forceFs = friction force

mg

Static friction• Fs,max = largest static friction

force possibleFs,max = s * Fg,y

• Block will not move if Fa,x ≤ Fs,max

• s – running shoe - loose gravel : 0.3– running shoe - grass: 1.5– bicycle racing tire concrete: 0.8

Fa,x

Fg,y

Fs

Fa,x = applied forceFs = friction force

mg

Dynamic friction

• Block moves if Fa,x > Fs,max (max static friction)

• Moving block experiences dynamic friction Fd = d * Fg,y

• d < s

Fa,x

Fg,y

Fs

Fa,x = applied forceFs = friction force

mg

Friction and Locomotion• Will a person’s shoe slip?• Will not slip if vector sum of Fg,x & Fg,z is less than

maximum static friction force.– Fg,x & Fg,z act parallel to surface

Fg,x

Fg,z

Fparallel = (Fg,x2 + Fg,z

2)0.5

Fparallel

When will shoe not slip?

• |Fparallel| ≤ Fs,max

– no slipping

• |Fparallel| ≤ s * Fg,y

• |Fparallel| / Fg,y ≤ s

Friction in walking & running• Constant velocity walking & running– |Fparallel| / Fg,y ≤ s

– Fg,z ~ 0

– |Fg,x| / Fg,y ≤ s

– Walk (1.2 m/s) & run (3.9 m/s): – |Fg,x| / Fg,y: 0.1 - 0.2

• µs for a running shoe– loose gravel: 0.3– dry grass: 1.5

Which will have the greater maximum static friction force?

A) AB) BC) The sameD) It depends

Ground reaction force during constant velocity cycling

• v = 12 m/s (26 mph): Fg,z ~ 0 N Fg,x = 25 N Fg,y = 500 N (half weight of bike & person)

Bicycle tire friction• No slip: Back wheel |Fg,x| /Fg,y ≤ µs,max

– |Fg,x|/ Fg,y = 25 N / 500 N = 0.05

• Static coefficient of friction for bicycle tire: – dry concrete: 0.8–wet concrete: 0.5– sand: 0.3

Forces parallel to the ground during acceleration in cycling

• Example: A person on a bicycle (total mass of 100 kg) is accelerating at 4 m/s2

– Acceleration force = Fg,x under back tireFg,x = ma = 100 kg * 4 m/s2 = 400 N

– |Fg,x|/ Fg,y = 400 N / 500 N = 0.8

• Static coefficient of friction for bicycle tire: – dry concrete: 0.8–wet concrete: 0.5– sand: 0.3

Friction in Object Manipulation

Friction & FBDs• Friction in locomotion allows there to be a

horizontal GRF• Don’t show GRF AND friction, just GRF

Problem: Friction force on slope

q

Fn

mg

Find maximum friction force in terms of mg, q, & µs.

Friction force on slope

Fs,max = Fn • µs

Fn = mg cos qFs,max = µs • mg cos qFparallel (force pulling downhill parallel to slope) = mg sin q

q

Fn

mg

Friction vs. Gravity force parallel

• m=70kg

• µs = 0.5• theta = 30 degrees• Solve for static friction force and the component

of gravitational force pulling parallel to the slope.• Will the block move? • Calculate its acceleration (µd = 0.4)